Đáp án:

\[x = \dfrac{1}{2};\,\,y = \dfrac{1}{2};\,\,z =  - \dfrac{1}{2}\]

Giải thích các bước giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\begin{array}{l}
\dfrac{x}{{y + z + 1}} = \dfrac{y}{{x + z + 1}} = \dfrac{z}{{x + y - 2}} = x + y + z\\
\dfrac{x}{{y + z + 1}} = \dfrac{y}{{x + z + 1}} = \dfrac{z}{{x + y - 2}} = \dfrac{{x + y + z}}{{\left( {y + z + 1} \right) + \left( {x + z + 1} \right) + \left( {x + y - 2} \right)}} = \dfrac{{x + y + z}}{{2.\left( {x + y + z} \right)}} = \dfrac{1}{2}\\
 \Rightarrow x + y + z = \dfrac{1}{2} \Rightarrow \left\{ \begin{array}{l}
y + z = \dfrac{1}{2} - x\\
x + z = \dfrac{1}{2} - y\\
x + y = \dfrac{1}{2} - z
\end{array} \right.\\
\dfrac{x}{{y + z + 1}} = \dfrac{y}{{x + z + 1}} = \dfrac{z}{{x + y - 2}} = \dfrac{1}{2}\\
 \Leftrightarrow \dfrac{x}{{\left( {\dfrac{1}{2} - x} \right) + 1}} = \dfrac{y}{{\left( {\dfrac{1}{2} - y} \right) + 1}} = \dfrac{z}{{\left( {\dfrac{1}{2} - z} \right) - 2}} = \dfrac{1}{2}\\
 \Leftrightarrow \dfrac{x}{{\dfrac{3}{2} - x}} = \dfrac{y}{{\dfrac{3}{2} - y}} = \dfrac{z}{{ - \dfrac{3}{2} - z}} = \dfrac{1}{2}\\
 \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{x}{{\dfrac{3}{2} - x}} = \dfrac{1}{2}\\
\dfrac{y}{{\dfrac{3}{2} - y}} = \dfrac{1}{2}\\
\dfrac{z}{{ - \dfrac{3}{2} - z}} = \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x = \dfrac{3}{2} - x\\
2y = \dfrac{3}{2} - y\\
2z =  - \dfrac{3}{2} - z
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3x = \dfrac{3}{2}\\
3y = \dfrac{3}{2}\\
3z =  - \dfrac{3}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y = \dfrac{1}{2}\\
z =  - \dfrac{1}{2}
\end{array} \right.
\end{array}\)

Vậy \(x = \dfrac{1}{2};\,\,y = \dfrac{1}{2};\,\,z =  - \dfrac{1}{2}\)